Proving 2^n < n without Equality: Why Ask?

[pivoxa15]

In my maths textbook it asks to prove 2^n<=n! for all n>=4

I could prove it no problems using induction but could show 2^n<n! without the equality inequality.

My question is why would the textbook ask for a weaker condition? Is it a misprint?

If I can show < then it automatically implies <= holds as well doesn't it?

 

[cristo]

If I can show < then it automatically implies <= holds as well doesn't it?

Yes, if x is strictly less than y, then x is certainly less than or equal to y. With regards to your former question, it might just be a typo- either way, you can solve the problem using induction.

 

[dextercioby]

It can't be a missprint. See what happens with n=4.

Daniel.

PS. Apparently, it can be a missprint.

 

[cristo]

It can't be a missprint. See what happens with n=4.

Daniel.

2^4=16 , 4!= 24, so 2^4<4!

 

[pivoxa15]

I was only asking the relationship between 2^n and n!

Looking at the graphs for 2^n and n! it seems that no where is 2^n=n!

n>=4 is definitely correct.

 

[Werg22]

Look at what are the factors of 2^n and the factors of n!...

 

[ssd]

There are two "=" signs in question. The first one is definitely a misprint but n>=4 is correct.